Question

In a certain​ region, women's heights are approximately normally distributed with mean 63.8 inches and standard deviation 4.2 inches. Men's heights are approximately normally distributed with mean 68.9 inches and standard deviation 4.5 inches.

a) At what doorway height would ​90% of men be able to pass through without having to​ duck?

b) What percentage of women would have to duck to pass through a doorway at the height found in part​ (a)?

Answer 1

a) The doorway height at which 90% of men can pass without having to duck is approximately 74.84 inches. b) About 0.6% of women would have to duck to pass through a doorway at the height found in part (a).

Step-by-step explanation:

a) To find the doorway height at which 90% of men can pass without having to duck, we need to find the z-score that corresponds to the 90th percentile. We can use the formula

z = (x - mean) / standard deviation.

Rearranging the formula, we have x = z * standard deviation + mean. Plugging in the values for men's heights, we get x = (1.2816 * 4.5) + 68.9 ≈ 74.84 inches.

Therefore, a doorway height of approximately 74.84 inches would allow 90% of men to pass through without ducking.

b) To determine the percentage of women that would have to duck to pass through a doorway at the height found in part (a), we need to find the percentage of women with heights less than the height found in part (a).

We can use the z-score formula to find the corresponding z-score. Plugging in the values for women's heights, we find the z-score to be (74.84 - 63.8) / 4.2 ≈ 2.6364.

Using a standard normal distribution table, we find that the percentage of women with heights less than 2.6364 standard deviations above the mean is approximately 99.4%.

Therefore, about 0.6% of women would have to duck to pass through a doorway at the height found in part (a).